Conservation of energy with a mass and pulley system.

[sdoi]

Homework Statement

Two boxes are attached to opposite ends of a rope passing over a frictionless pulley as shown below. The mass of Box A is 15kg and the mass of box B is 12kg. The system is originally at rest with the bottom of box A at a height of o.85m above the floor. When the system is released, the boxes will move. Use conservation of energy to determine the speed with which Box A will contact the floor.

Homework Equations

Eg=mgΔh
Ek= 1/2 mv^2/2
ƩFy=may

The Attempt at a Solution

I started off by drawing free body diagrams of each mass, one at rest, and one in motion.

For mass A:
at rest,
ƩFy=0
Ft+Eg=0
Ft= Eg
= mgΔh
=(15kg)(9.8m/s^2)(0.85m)
Ft=125N

in motion,
ƩFy= may
Fg(A)-Ft= m(A)ay

For mass B:
at rest,
ƩFy=0
Fn-mg=0
Fn=mg
=(12kg)(9.8m/s^2)
Fn=117.6N

in motion,
ƩFy=may
Ft-Fg(B)= m(B)ay

I'm not sure if my original statement of Ft=Eg is accurate... and from this point on I don't know where to go.

 

[grzz]

Why not find the initial and final energies and equate them?

 

[sdoi]

As in, Etotal= Eg, Etotal'=Ek ?

 

[grzz]

System is originally at rest. Hence initial energy = ...
Final energy = ...

 

[asteriatic]

Great job on drawing the free body diagrams and identifying the forces acting on each mass. Your initial statement of Ft=Eg is correct, as they are equal and opposite forces in this system. From here, you can use the conservation of energy principle to solve for the speed with which Box A will contact the floor. This principle states that the total energy of a closed system remains constant, and in this case, the system is only experiencing gravitational potential energy (Eg) and kinetic energy (Ek). Therefore, we can equate the initial potential energy (Eg) to the final kinetic energy (Ek):

Eg = Ek
mgΔh = 1/2 mv^2

Solving for v, we get:

v = √(2gh)

Plugging in the values given in the problem, we get:

v = √(2*9.8m/s^2*0.85m) = 4.38 m/s

Therefore, the speed with which Box A will contact the floor is 4.38 m/s. This solution is only valid if the rope remains taut throughout the motion, and if the pulley is frictionless. Great job on applying the conservation of energy principle to this mass and pulley system!